[tex] \frac{1}{x}- \frac{1}{x+1}= \frac{(x+1)-x}{x(x+1)}= \frac{1}{x(x+1)} \leq 0[/tex]
Therefore x(x+1)≤0, and if two numbers multiply to a negative number, one of them has to be negative and the other has to be positive. In this case, because x+1>x, x≤0, and x+1≥0
Therefore -1≤x≤0, and x can take up the value of any real number in the interval [-1,0]
